EE 333, Communication Networks Mid-Term Exam (2014-15S)

Maximum Marks = 30 Time = 2 hours

1. For N-packet blocks, let XN be the time taken from the instant the transmission of the block begins for the first time to the time when the transmission of the next block can be started.

(a) N=1

1 1

1

1

11

2 2 (1 )

1 22

where1 2

X T d T d X

dXT d dTX bT b T

[2]

(b) N=2

2 22 2

2

1

2

2(2 2 ) 2 2 2 (1 ) 2 2 (1 )

(2 ) (2 2 ) 2( 2 )(1 ) 2 (2 ) 2 (3 2 )

2 2 (3 2 ) 1 (3 2 )

(2 ) (2 )

T dX T d T d T d X

X T d T d T d

T d bX

[4]

(c) N=3 with d=0 10 20

2T TX X

3 2 2 330 10 20 30

3 230

2 3 230

2 230

30

30

3 3 3 (1 ) 3 3 (1 ) 3 (1 )

1 (1 ) 3 3 (1 ) 6(1 )

3 3 3 1 (1 ) 2(1 )

3 3 3 3 3

3 3

X T T X T X T X

X T T T

X T

X T

T TX

X

[4]

We can logically argue that in this ARQ scheme, each packet is sent α-1 times as in a simple Stop-&-Wait ARQ

system. (This is because every packet is repeated until it is correctly received and only the packets in error are

repeated.) The differences between schemes with different block-lengths arise only because of the effects of the

propagation delay d. When d=0, all the schemes will perform similarly and will have the same efficiency as the

equivalent Stop-&-Wait approach.

2. (a) Consider the code word to be 1 2

1 2 1 0( ) ...........n n

n nT x T x T x T x T

A circular right shift of T(x) by one position yields the polynomial T*(x) as follows –

1 2 3

0 1 2 2 1*( ) ...........n n n

n nT x T x T x T x T x T

Note that T*(x) can also be obtained by circularly left shifting T(x) by n-1 positions. We can show (given below) that circularly left shifting T(x) by one position also creates a code word. Therefore, circularly left shifting by n-1 positions will also create a code word. To show that circularly left shifting T(x) by one position creates another code word -

Circular Left shift by one position, 1 1 1

ˆ( ) ( ) ( ) (1 )n n

n n nT x xT x T x T xT x T x

In ˆ( )T x , (a) T(x) is obviously divisible by g(x)

(b) (1+xn) has (1+x) as a factor and (1+xn) is also divisible by P(x). Therefore (1+xn) is also divisible by g(x)

Therefore ˆ( )T x is also a code word.

(b) Comparing ( )C x and ( )C x , it is easy to see that the error polynomial ( )E x is –

10 9 8 4 2( ) 1E x x x x x x

This error pattern will be detectable if ( )E x is divisible by ( )g x . We can check this by long division

and verify that Rem10 9 8 4 2

8 2

10

1

x x x x x

x x x

Note: Actually 10 9 8 4 2 2 8 21 ( 1)( 1)x x x x x x x x x x from the long division

Therefore, the error pattern ( )E x will not be detectable by CRC-8 and ( )C x will be accepted by the

receiver as a valid code word. We can also verify that

17 14 12 11 10 9 7 5 4 2

9 6 4 8 2

( ) 1

( 1)( 1)

C x x x x x x x x x x x x

x x x x x x

Therefore ( )C x is indeed a valid code word and will be accepted by the receiver.

[By long division, we get

17 14 12 11 8 7 5 9 6 4 2 8 2( ) ( )( 1)C x x x x x x x x x x x x x x x x x ]

3. (a) State Transition Diagram

(b) 3 2 2 2 2 2 2

0 1 1 23 (1 ) (1 ) (1 ) 2 (1 ) 2 (1 )P q q q P q q P q q q q q P q q

Therefore, 2 2

1 0 2 1 02 2 4

(3 2 ) 1 (3 2 )(1 )

(1 ) 2(1 ) 2(1 )

q q q q q q q qP P P P P

q q q

(c) It is easy to show that 2 2

0 1 2 33 (1 ) 3 (1 )S q q P P P P q q

Differentiating this with respect to q, we find that the maximum occurs at 1

3q and the

maximum value of S is max

40.444

9S